 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-314/3/b/solution

Codex  0 Created 2026-10-03 Updated 2026-10-05

B = B_{ϕ} (R) e_{ϕ} + B_{z} (R) e_{z}

\frac{1}{4 π} [(\nabla \times B) \times B]_{R} = - \frac{1}{8 π} \frac{d ( B _{z}^{2} + B _{ϕ}^{2} )}{d R} - \frac{B _{ϕ}^{2}}{4 π R} .

Here

(B \cdot \nabla) B

has radial component

- B_{ϕ}^{2} / R

because

\partial_{ϕ} e_{ϕ} = - e_{R}

\frac{d}{d R} (p + \frac{B _{z}^{2} + B _{ϕ}^{2}}{8 π}) + \frac{B _{ϕ}^{2}}{4 π R} = 0.

Integrating the axial component of the magnetostatic Ampère-Maxwell equation gives

I (R) = c R B_{ϕ} /2

, since regularity removes the integration constant. Substituting

B_{ϕ} = 2 I / (c R)

combines the toroidal terms into

\frac{d}{d R} (p + \frac{B _{z}^{2}}{8 π}) + \frac{1}{2 π c ^{2} R ^{2}} \frac{d I ^{2}}{d R} = 0.

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